Solving Quadratic Inequalities – Example 1.5.1 | 解二次不等式

📚 Solving Quadratic Inequalities – Example 1.5.1 | 解二次不等式 – 例 1.5.1

This article walks through Example 1.5.1, a typical AQA A-Level Mathematics question on quadratic inequalities. Quadratic inequalities are a core topic in Pure Mathematics, appearing in Paper 1 and often forming the foundation for later work on functions, graphs, and optimisation.

本文带您逐步解析例 1.5.1——一道典型的 AQA A-Level 数学二次不等式题目。二次不等式是纯数学部分的核心内容，常见于 Paper 1，并常常为后续的函数、图像和最优化问题奠定基础。

1. Understanding Quadratic Inequalities | 理解二次不等式

A quadratic inequality is an inequality that involves a quadratic expression — that is, an expression of the form ax² + bx + c, where a ≠ 0. The inequality can take one of four forms: f(x) > 0, f(x) < 0, f(x) ≥ 0, or f(x) ≤ 0.

二次不等式是涉及二次表达式的不等式，即形如 ax² + bx + c（其中 a ≠ 0）的表达式。不等式可以是以下四种形式之一：f(x) > 0、f(x) < 0、f(x) ≥ 0 或 f(x) ≤ 0。

Unlike linear inequalities, which have a single interval of solutions, quadratic inequalities generally produce either two separate intervals or a single bounded interval. The key to solving them is to first find the roots of the corresponding quadratic equation.

与只有一个解区间的线性不等式不同，二次不等式通常产生两个分离的区间或一个有限区间。解题的关键是首先求出对应二次方程的根。

2. The Role of the Discriminant | 判别式的作用

The discriminant, Δ = b² − 4ac, tells us whether the quadratic equation ax² + bx + c = 0 has two distinct real roots (Δ > 0), one repeated real root (Δ = 0), or no real roots (Δ < 0).

判别式 Δ = b² − 4ac 告诉我们二次方程 ax² + bx + c = 0 有两个不同的实根（Δ > 0）、一个重实根（Δ = 0）还是没有实根（Δ < 0）。

For a quadratic inequality, knowing the number of roots tells us how the parabola crosses the x-axis, which directly determines the structure of the solution set. If Δ < 0 and a > 0, the expression is always positive; if Δ < 0 and a < 0, it is always negative.

对于二次不等式，了解根的个数可以告诉我们抛物线如何穿过 x 轴，这直接决定了解集的结构。如果 Δ < 0 且 a > 0，则表达式恒为正；如果 Δ < 0 且 a < 0，则恒为负。

3. The Critical Points Approach | 临界点法

The most reliable method for solving a quadratic inequality is the critical points method. First, rewrite the inequality in the form f(x) = ax² + bx + c compared with zero. Then factorise f(x) if possible, or use the quadratic formula to find the roots.

解二次不等式最可靠的方法是临界点法。首先将不等式改写为 f(x) = ax² + bx + c 与零比较的形式。然后对 f(x) 进行因式分解（如果可能的话），或使用二次方程求根公式来求根。

These roots divide the real number line into distinct intervals. On each interval, the sign of f(x) is constant. By testing one point from each interval, we can determine whether that interval belongs to the solution set.

这些根将实数轴划分为若干个区间。在每个区间内，f(x) 的符号恒定。通过在每个区间测试一个点，我们就可以判断该区间是否属于解集。

Step 1: Solve f(x) = 0 → find critical points x₁ and x₂

步骤 1：解 f(x) = 0 → 求出临界点 x₁ 和 x₂

4. Sign Testing on Intervals | 区间符号检验

Once the critical points are found, we draw a number line and mark the roots. Suppose the roots are x₁ and x₂, with x₁ < x₂. The number line is then split into three intervals: x < x₁, x₁ < x < x₂, and x > x₂.

一旦求出临界点，我们画一条数轴并标出这些根。假设根为 x₁ 和 x₂，且 x₁ < x₂。那么数轴被分为三个区间：x < x₁、x₁ < x < x₂ 和 x > x₂。

To determine the sign of f(x) on each interval, select a convenient test value — for example, x = 0 if it is not a root — and substitute it into the expression. Record whether the result is positive or negative.

为了确定 f(x) 在每个区间上的符号，选取一个方便的测试值——例如 x = 0（如果它不是根）——并将其代入表达式中。记录结果为正还是为负。

For a quadratic with a > 0, the parabola opens upward, so f(x) is positive outside the roots and negative between the roots. For a < 0, the signs are reversed. However, relying on test values is always safe and avoids sign errors.

对于 a > 0 的二次函数，抛物线向上开口，因此 f(x) 在两根之外为正，在两根之间为负。对于 a < 0，符号相反。然而，依赖测试值总是安全的，可以避免符号错误。

5. Example 1.5.1 – Worked Solution | 例 1.5.1 – 详细解答

Problem | 题目： Solve the inequality x² − 5x + 6 > 0. | 解不等式 x² − 5x + 6 > 0。

We begin by solving the corresponding equation x² − 5x + 6 = 0. This quadratic factorises neatly as (x − 2)(x − 3) = 0, giving roots x = 2 and x = 3.

首先解对应的方程 x² − 5x + 6 = 0。这个二次式可以简洁地因式分解为 (x − 2)(x − 3) = 0，得到根 x = 2 和 x = 3。

These two roots divide the number line into three intervals: (−∞, 2), (2, 3), and (3, ∞). We now test each interval.

这两个根将数轴分为三个区间：(−∞, 2)、(2, 3) 和 (3, ∞)。现在我们逐个区间进行检验。

For the interval x < 2, choose x = 0. Substituting, we get 0² − 5(0) + 6 = 6 > 0, so this interval satisfies the inequality.

对于 x < 2 的区间，选取 x = 0。代入得 0² − 5(0) + 6 = 6 > 0，因此该区间满足不等式。

For the interval 2 < x < 3, choose x = 2.5. Substituting, we get (2.5)² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25 < 0, so this interval does not satisfy the inequality.

对于 2 < x < 3 的区间，选取 x = 2.5。代入得 (2.5)² − 5(2.5) + 6 = 6.25 − 12.5 + 6 = −0.25 < 0，因此该区间不满足不等式。

For the interval x > 3, choose x = 4. Substituting, we get 4² − 5(4) + 6 = 16 − 20 + 6 = 2 > 0, so this interval satisfies the inequality.

对于 x > 3 的区间，选取 x = 4。代入得 4² − 5(4) + 6 = 16 − 20 + 6 = 2 > 0，因此该区间满足不等式。

Solution: x < 2 or x > 3 | 解：x < 2 或 x > 3

6. Graphical Interpretation | 图像解读

The parabola y = x² − 5x + 6 opens upward because the coefficient of x² is positive. It crosses the x-axis at x = 2 and x = 3. The portion of the curve above the x-axis corresponds to y > 0, which occurs when x is to the left of 2 or to the right of 3.

抛物线 y = x² − 5x + 6 向上开口，因为 x² 的系数为正。它在 x = 2 和 x = 3 处穿过 x 轴。曲线位于 x 轴上方的部分对应 y > 0，即当 x 位于 2 的左侧或 3 的右侧时。

This graphical view confirms that the solution to x² − 5x + 6 > 0 is indeed x < 2 or x > 3. It also helps visualise why the middle interval is excluded: between the two roots, the curve dips below the x-axis and the function takes negative values.

从图像角度看，这确认了 x² − 5x + 6 > 0 的解确实是 x < 2 或 x > 3。它还有助于直观地理解为什么中间的区间被排除：在两个根之间，曲线降到 x 轴以下，函数取负值。

When the inequality involves ≥ or ≤, the roots themselves are included. In this example, since the inequality is strict (>), the critical points x = 2 and x = 3 are not part of the solution set.

当不等式涉及 ≥ 或 ≤ 时，根本身包含在解集中。在本题中，由于不等式是严格的（>），临界点 x = 2 和 x = 3 不属于解集。

7. Interval Notation | 区间表示法

The solution x < 2 or x > 3 can be written in set notation as {x : x < 2} ∪ {x : x > 3}. In interval notation, this is written as (−∞, 2) ∪ (3, ∞).

解 x < 2 或 x > 3 可以用集合符号写为 {x : x < 2} ∪ {x : x > 3}。在区间表示法中，这写作 (−∞, 2) ∪ (3, ∞)。

The round brackets indicate that the endpoints 2 and 3 are excluded. If the inequality had been x² − 5x + 6 ≥ 0, the solution would be (−∞, 2] ∪ [3, ∞), using square brackets to include the endpoints.

圆括号表示端点 2 和 3 被排除。如果不等式是 x² − 5x + 6 ≥ 0，则解为 (−∞, 2] ∪ [3, ∞)，用方括号将端点包含在内。

Proper use of interval notation is essential for AQA A-Level questions, as marks are often awarded for the final presentation of the solution set, not just for the numerical answer.

正确使用区间表示法对于 AQA A-Level 题目至关重要，因为分数通常不仅取决于数值答案，还取决于解集的最终表述形式。

8. Common Mistakes to Avoid | 常见错误与防范

Incorrectly reversing the inequality sign: When multiplying or dividing by a negative number, the inequality direction flips. In quadratic inequalities, this can happen when multiplying by −1 to change the sign of all terms.

错误地反转不等式符号：当乘以或除以一个负数时，不等号方向会反转。在二次不等式中，当乘以 −1 来改变所有项的符号时可能发生这种情况。
Imposing the wrong interval: A common mistake is to write the solution of f(x) > 0 as x₁ < x < x₂ instead of x < x₁ or x > x₂. Always check the direction of the parabola or test a point.

选择了错误的区间：一个常见错误是将 f(x) > 0 的解写为 x₁ < x < x₂，而不是 x < x₁ 或 x > x₂。始终检查抛物线的开口方向或测试一个点。
Forgetting to check the discriminant: If Δ < 0, the solution is either all real numbers or none, depending on the sign of a. Skipping this check can lead to missing the full picture.

忘记检查判别式：如果 Δ < 0，则解要么是全体实数，要么是空集，取决于 a 的符号。跳过此检查可能导致遗漏整体情况。

9. When the Quadratic Cannot Be Factorised | 当二次式无法因式分解时

If the quadratic cannot be factorised easily, we can use the quadratic formula to find the roots:

如果二次式不易因式分解，我们可以使用求根公式来求出根：

x = (−b ± √(b² − 4ac)) / 2a

Once the roots are found, the critical points method proceeds exactly as before. The roots may be irrational, but the sign-testing procedure remains unchanged.

求出根后，临界点法的步骤与之前完全相同。根可能是无理数，但符号检验的过程不变。

For example, to solve x² + 2x − 2 ≥ 0, we compute the discriminant Δ = 4 + 8 = 12, so the roots are x = −1 ± √3. The solution is then x ≤ −1 − √3 or x ≥ −1 + √3.

例如，要解 x² + 2x − 2 ≥ 0，我们计算判别式 Δ = 4 + 8 = 12，因此根为 x = −1 ± √3。解为 x ≤ −1 − √3 或 x ≥ −1 + √3。

10. Special Case: Repeated Roots | 特殊情况：重根

When the discriminant equals zero, the quadratic has one repeated root and the graph touches the x-axis at exactly one point. In this case, the expression f(x) is always non-negative (if a > 0) or always non-positive (if a < 0), except at the root.

当判别式等于零时，二次方程有一个重根，图像仅在一点处接触 x 轴。在这种情况下，表达式 f(x) 始终为非负（若 a > 0）或始终为非正（若 a < 0），除了在根这一点处。

Consider x² − 4x + 4 ≥ 0. Since x² − 4x + 4 = (x − 2)² ≥ 0 for all real x, the solution is all real numbers. On the other hand, x² − 4x + 4 > 0 gives all real numbers except x = 2.

考虑 x² − 4x + 4 ≥ 0。由于 x² − 4x + 4 = (x − 2)² ≥ 0 对一切实数 x 成立，因此解为全体实数。另一方面，x² − 4x + 4 > 0 的解为除 x = 2 以外的所有实数。

This special case is often tested in AQA papers because it combines algebraic manipulation with a precise understanding of strict versus non-strict inequalities.

这种特殊情况在 AQA 试卷中经常考查，因为它结合了代数运算以及对严格不等式与非严格不等式的精确理解。

11. Practice Similar Example | 同类练习

Try this | 试做这道题： Solve the inequality 2x² + 5x − 3 ≤ 0. | 解不等式 2x² + 5x − 3 ≤ 0。

We first factorise: 2x² + 5x − 3 = (2x − 1)(x + 3). Setting this equal to zero gives x = 1/2 and x = −3.

首先进行因式分解：2x² + 5x − 3 = (2x − 1)(x + 3)。令其等于零，得到 x = 1/2 和 x = −3。

Since the leading coefficient is positive, the parabola opens upward. The expression is negative or zero between the roots, so the solution to the inequality is −3 ≤ x ≤ 1/2.

由于首项系数为正，抛物线向上开口。表达式在两根之间为负或零，因此不等式的解为 −3 ≤ x ≤ 1/2。

Answer: −3 ≤ x ≤ 1/2 | 答案：−3 ≤ x ≤ 1/2

12. Summary of Steps | 步骤总结

Step \| 步骤	Action \| 操作
1	Compare the quadratic to zero: make sure one side is 0. \| 将二次式与零比较：确保一边为 0。
2	Solve the corresponding quadratic equation by factorisation or the quadratic formula. \| 通过因式分解或求根公式解对应的二次方程。
3	Mark the roots on a number line to create intervals. \| 在数轴上标出根，形成区间。
4	Test one value from each interval to determine the sign of f(x). \| 从每个区间选取一个测试值，确定 f(x) 的符号。
5	Write the solution in set or interval notation, respecting strict vs non-strict inequalities. \| 用集合或区间表示法写出解集，注意严格不等式与非严格不等式的区别。

A consistent, methodical approach to quadratic inequalities will serve you well across many topics in A-Level Mathematics. Once mastered, this skill transfers directly to solving polynomial inequalities, rational inequalities, and optimisation problems in calculus.

以一致且系统的方法处理二次不等式，将为您在 A-Level 数学的许多主题中打下坚实基础。一旦掌握，这一技能可以直接迁移到解多项式不等式、有理不等式以及微积分中的最优化问题。

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Step \| 步骤	Action \| 操作
1	Compare the quadratic to zero: make sure one side is 0. \| 将二次式与零比较：确保一边为 0。
2	Solve the corresponding quadratic equation by factorisation or the quadratic formula. \| 通过因式分解或求根公式解对应的二次方程。
3	Mark the roots on a number line to create intervals. \| 在数轴上标出根，形成区间。
4	Test one value from each interval to determine the sign of f(x). \| 从每个区间选取一个测试值，确定 f(x) 的符号。
5	Write the solution in set or interval notation, respecting strict vs non-strict inequalities. \| 用集合或区间表示法写出解集，注意严格不等式与非严格不等式的区别。

Solving Quadratic Inequalities – Example 1.5.1 | 解二次不等式 – 例 1.5.1